__Question 3.8__

On a two-lane road, car A is travelling with a speed of 36 km h

^{-1}. Two cars B and C approach car A in opposite directions with a speed of 54 km h

^{-1}each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident ?

__Solution:__

Velocity of car A,

*v*

_{A}= 36 km/h = 10 m/s

Velocity of car B,

*v*

_{B}= 54 km/h = 15 m/s

Velocity of car C,

*v*

_{C}= 54 km/h = 15 m/s

Relative velocity of car B with respect to car A,

*v*

_{BA}=

*v*

_{B}–

*v*

_{A}

= 15 – 10 = 5 m/s

Relative velocity of car C with respect to car A,

*v*

_{CA}=

*v*

_{C }– (–

*v*

_{A})

= 15 + 10 = 25 m/s

At a certain instance, both cars B and C are at the same distance from car A i.e.,

*s*= 1 km = 1000 m

Time taken (

*t*) by car C to cover 1000 m = 1000 / 25 = 40 s

Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s.

From second equation of motion, minimum acceleration (

*a*) produced by car B can be obtained as:

s = ut + (1/2)at

^{2}

1000 = 5 X 40 + (1/2) X a X (40)

^{2}

a = 1600 / 1600 = 1 ms

^{-2}

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